zs

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

受欢迎的三角函数 >

证明 6cos^2(A)-5=1-6sin^2(A)

解答

证明

6 cos^{2} (A) - 5 = 1 - 6 sin^{2} (A)

解答

真

求解步骤

6 cos^{2} (A) - 5 = 1 - 6 sin^{2} (A)

调整左侧

6 cos^{2} (A) - 5

使用三角恒等式改写

6 cos^{2} (A) - 5

使用毕达哥拉斯恒等式:

cos^{2} (x) + sin^{2} (x) = 1

cos^{2} (x) = 1 - sin^{2} (x)

= 6 (1 - sin^{2} (A)) - 5

化简

6 (1 - sin^{2} (A)) - 5 : - 6 sin^{2} (A) + 1

6 (1 - sin^{2} (A)) - 5

乘开

6 (1 - sin^{2} (A)) : 6 - 6 sin^{2} (A)

6 (1 - sin^{2} (A))

使用分配律:

a (b - c) = ab - a c

a = 6, b = 1, c = sin^{2} (A)

= 6 \cdot 1 - 6 sin^{2} (A)

数字相乘:

6 \cdot 1 = 6

= 6 - 6 sin^{2} (A)

= 6 - 6 sin^{2} (A) - 5

化简

6 - 6 sin^{2} (A) - 5 : - 6 sin^{2} (A) + 1

6 - 6 sin^{2} (A) - 5

对同类项分组

= - 6 sin^{2} (A) + 6 - 5

数字相加/相减:

6 - 5 = 1

= - 6 sin^{2} (A) + 1

= - 6 sin^{2} (A) + 1

= - 6 sin^{2} (A) + 1

= - 6 sin^{2} (A) + 1

= 1 - 6 sin^{2} (A)

我们已展示，在两侧可以有相同的形式

\Rightarrow 真

流行的例子

证明 (sin(2θ))/(sin(θ))=2

p ro v e \frac{sin ( 2 θ )}{sin ( θ )} = 2

证明 cos^2(x)-sin^2(x)+1=2cos^2(x)

p ro v e cos^{2} (x) - sin^{2} (x) + 1 = 2 cos^{2} (x)

证明-2sin(2x+pi)=-2sin(2x-pi)

p ro v e - 2 sin (2 x + π) = - 2 sin (2 x - π)

证明 2sin(a+b)sin(a-b)=cos(2b)-cos(2a)

p ro v e 2 sin (a + b) sin (a - b) = cos (2 b) - cos (2 a)

证明 1/(sin(x))=2

p ro v e \frac{1}{sin ( x )} = 2